# Find N in a Limit of a Sequence Homework

• kwal0203
In summary: So, for example:##|\frac{n^2 + 3n +1}{n^3 - 2n^2 + 6n - 1}|##.Now ##|n^2 + 3n + 1| < |n^3 + 3n^2 + 3n| = |n^3| < |n^3 - 2n^2 + 6n - 1|##. So:##|n^2 + 3n + 1| < |n^3 - 2n^2 + 6n - 1| < |n^3| + |2n^2| + |6n| + |
kwal0203

## Homework Statement

Suppose $$a_{n}=\frac{n^2-2n+1}{2n^2+4n-1}$$

For each positive number $$\epsilon$$, find a number N such that:

$$\mid a_{n} - L\mid < \epsilon$$ whenever n > N.

## The Attempt at a Solution

$$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon$$

$$\mid \frac{3-8n} {4n^2+8n-2} \mid < \epsilon$$

Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

Thanks!

Last edited:
kwal0203 said:

## Homework Statement

Suppose $$a_{n}=\frac{n^2-2n+1}{2n^2+4n-1}$$

For each positive number \epsilon, find a number N such that:

\mid a_{n} - L\mid < \epsilon whenever n > N.

## The Attempt at a Solution

\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon

\mid \frac{3-8n} {2n^2+4n-1} \mid < \epsilon

Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

Thanks!

Try some estimating.

PeroK said:
Try some estimating.

Fixed it up.

Estimating what exactly?

kwal0203 said:
Fixed it up.

Estimating what exactly?

Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

Show that ##5n^2 + 3n + 133## tends to ##\infty## an ##n \rightarrow \infty##

Estimating technique:

For ##n > 1## we have ##n^2 > n## hence ##5n^2 > n##. So:

##5n^2 + 3n + 133 > 4n + 133 > 4n > n##

Now, we can use a simple ##N## in the limit argument. Whereas, trying to find ##N## for the orginal quadratic would be messy.

kwal0203 said:
$$\mid \frac{3-8n} {2n^2+4n-1} \mid < \epsilon$$

PS I think you've lost a factor of ##2## from the denominantor.

PeroK said:
PS I think you've lost a factor of ##2## from the denominantor.

Yes I definitely have, thanks.

PeroK said:
Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

Show that ##5n^2 + 3n + 133## tends to ##\infty## an ##n \rightarrow \infty##

Estimating technique:

For ##n > 1## we have ##n^2 > n## hence ##5n^2 > n##. So:

##5n^2 + 3n + 133 > 4n + 133 > 4n > n##

Now, we can use a simple ##N## in the limit argument. Whereas, trying to find ##N## for the orginal quadratic would be messy.

So it goes kinda likes this:

For n > 1 we have $$n^2>n$$ hence, $$2n^2>n$$ So:

$$2n^2+4n-1>5n-1>5n>n$$

And in a similar way:

For n > 1 we have $$3 - 8n > 3 - 8 = -5$$ So:

$$\mid \frac{-5}{n} \mid < \epsilon$$

and,

$$n > 5 / \epsilon$$

Is that what you mean?

kwal0203 said:
So it goes kinda likes this:

For n > 1 we have $$n^2>n$$ hence, $$2n^2>n$$ So:

$$2n^2+4n-1>5n-1>5n>n$$

And in a similar way:

For n > 1 we have $$3 - 8n > 3 - 8 = -5$$ So:

$$\mid \frac{-5}{n} \mid < \epsilon$$

and,

$$n > 5 / \epsilon$$

Is that what you mean?

First, you need to take that ##\epsilon## out of your working. That seems to be a common mistake to introduce ##\epsilon## into the inequality too soon. Second, your estimates have gone a bit wonky!

Let me show you what I mean. You should leave out the ##\epsilon## for now and do:

##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots##

Note that the numerator is less than ##8n## and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because ##8n/n## doesn't converge to 0. So, you need an estimate based on ##n^2## in this case.

It's only when you've finished this algebra that you bring in ##\epsilon##. For example:

Let ##\epsilon > 0 ## and take ##N > 1/ \epsilon## (or whatever you think you need from your calculations). Then ##n > N## implies:

##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon##

PeroK said:
First, you need to take that ##\epsilon## out of your working. That seems to be a common mistake to introduce ##\epsilon## into the inequality too soon. Second, your estimates have gone a bit wonky!

Let me show you what I mean. You should leave out the ##\epsilon## for now and do:

##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots##

Note that the numerator is less than ##8n## and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because ##8n/n## doesn't converge to 0. So, you need an estimate based on ##n^2## in this case.

It's only when you've finished this algebra that you bring in ##\epsilon##. For example:

Let ##\epsilon > 0 ## and take ##N > 1/ \epsilon## (or whatever you think you need from your calculations). Then ##n > N## implies:

##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon##

Ok I think I get it now. Thanks for your help I'll do some more work on it.

kwal0203 said:
Ok I think I get it now. Thanks for your help I'll do some more work on it.

The simplest way to explain estimating is perhaps:

##|\frac{a(n)}{b(n)}| < |\frac{c(n)}{d(n)}|##

Where ##|a(n)| < |c(n)|## and ##|b(n)| > |d(n)|## and ##c(n), d(n)## are some simple expressions in ##n## that reduce to something that can easily be shown to have a limit of ##0##.

kwal0203

## 1. What is the purpose of finding N in a limit of a sequence?

The purpose of finding N in a limit of a sequence is to determine the value of N that will make the sequence approach a specific limit. This is important in understanding the behavior of a sequence and its convergence or divergence.

## 2. How do you find N in a limit of a sequence?

To find N in a limit of a sequence, you need to first identify the sequence and its limit. Then, you can use algebraic manipulation or a specific formula such as the epsilon-delta definition of a limit to solve for N.

## 3. What are the common methods used to find N in a limit of a sequence?

The most common methods used to find N in a limit of a sequence are the epsilon-delta method, the squeeze theorem, and the geometric series formula. These methods involve either algebraic manipulation or using specific mathematical formulas.

## 4. Is it possible to find multiple values of N for a given limit of a sequence?

Yes, it is possible to find multiple values of N for a given limit of a sequence. This is because there can be different approaches or methods to finding N, and each method may yield a different value. It is important to check for consistency and accuracy when finding N through different methods.

## 5. How can finding N in a limit of a sequence be applied in real-life situations?

Finding N in a limit of a sequence can be applied in various real-life situations, such as in engineering, physics, and economics. For example, it can be used to model the growth of a population or the decay of a radioactive substance. It can also be used to optimize processes and systems by determining the optimal value of N for a given limit.

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